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Applied Statistics and Probability for Engineers, 7th edition 2017

5-18

Section 5-5

5.5.1 a) percentage of slabs classified as high with probability p1 = 0.05

b) X is the number of voids independently classified as high X 0

5.5.2 a) Probability for the kth landing page = pk = 0.25

Applied Statistics and Probability for Engineers, 7th edition 2017

5-19

5.5.3 Because

= 0 and X and Y are normally distributed, X and Y are independent. Therefore,

5.5.4 a)

Applied Statistics and Probability for Engineers, 7th edition 2017

5-20

b)

c)

5.5.5 Because

= 0 and X and Y are normally distributed, X and Y are independent. Therefore,

X = 0.1 mm,

X = 0.00031 mm,

Y = 0.23 mm,

Y = 0.00017 mm

5.5.6 a)

= cov(X,Y)/

x

y = 0.6, cov(X,Y) = 0.6(2)(5) = 6

Applied Statistics and Probability for Engineers, 7th edition 2017

5.5.7

−

−

−+

− − − −

==

2

2()

()

1

2 2 2

1

( , ) 2

Y

X

XY

y

x

XY

XY

f x y dxdy e dxdy

5.5.8 a) X1, X2 and X3 are binomial random variables when considered individually, that is, the marginal

probability distributions are binomial. Their joint distribution is multinomial.

= = =

1 1 2 2 3 3

( , , )

P X x X x X x

Section 5-6

5.6.1 a) E(3X + 2Y) = 3(2) + 2(6) = 18

5.6.3 a) X N(0.1, 0.00031) and Y N(0.23, 0.00017) Let T denote the total thickness.

5.6.4 a) X: time of wheel throwing. X ~ N(40,4)

5.6.5 X = time of ACL reconstruction surgery for high-volume hospitals.

5.6.6 a) Let

X

denote the average fill-volume of 100 cans

==

2

0.5 100 0.05

X

5.6.7 X ~ N(160, 900)

5.6.8 D = A − B − C

5-23

5.6.9

12

( ) (2 2 )

V Y V X X

=+

5.6.10 Let Y be the rate of return for the entire investment after one year,

5.6.11 Let Xi be the demand in month i.

a)

1 2 12

...

Z X X X

= + + +

Applied Statistics and Probability for Engineers, 7th edition 2017

5-24

Section 5-7

5.7.1

1

() 4

Y

fy=

at y = 3, 5, 7, 9

−

−

1

2

1

y

e

5.7.4 a) Now,

2

0

bv

av e dv

−

must equal one. Let u = bv, then

2

2

3

00

1uu

udu a

a e u e du

b

bb

−−

==

. From the

5.7.5 If y = ex, then x = ln y for 1 ≤ x ≤ 2 and e1 ≤ y ≤ e2. Thus,

11

( ) (ln )

YX

f y f y yy

==

for 1 ≤ ln y ≤ 2.

5.7.6

ln

W

Ye

WY

=

=

5.7.7

( )

= + + − 0.5

22

1 2 1 2 1 2

(cos cos )rrrrr

Applied Statistics and Probability for Engineers, 7th edition 2017

5-25

Then, f(r) = f(

2)|J|, where

5.7.8 Here, X is lognormal with

= 5.2933, and

2 = 0.00995

Y = X4

Section 5-8

5.8.1 a)

−−

+

= = =

−

= = = = = −

11

( 1)

1 0 0

1 1 (1 )

( ) ( ) (1 )

t t tm

m m m

tX tx t x tx

Xt

x x x

e e e

M t E e e e e

m m m me

Applied Statistics and Probability for Engineers, 7th edition 2017

5-26

5.8.3 a)

−

==

= = − = −

−

1

11

( ) ( ) (1 ) ( (1 ))

1

tX tx x t x

X

xx

p

M t E e e p p e p

p

b)

( ) ( )

( )

=

=

− − − − −

= = = = −−

12

0

0

1 (1 ) (1 )

()

( ) '

1 (1 )

t t t t

X

t

t

t

pe p e pe p e

dM t

EX dt pe

5.8.5 a)

−−

==

= = =

2 ( 2)

00

( ) ( ) (4 ) 4

tX tx x t x

X

xx

M t E e e xe dx xe dx

Applied Statistics and Probability for Engineers, 7th edition 2017

5-27

5.8.6

a)

=

=

−

= = = = − =

− − − −

1 1 1

( ) ( ) ()

x

tx t t t t

tX tx

X

x

e e e e e

M t E e e dx t t t t

( ) ( )

→

− − − + −

=−

2 2 2

23

0

2 2 2

' lim ()

t t t t t t

t

t e e t e e e e

t

5.8.7 a)

−−

= = =

()

( ) ( )

tX tx x t x

X

M t E e e e dx e dx

Applied Statistics and Probability for Engineers, 7th edition 2017

5-28

5.8.8 a)

( )

−− − −

= = =

11 ( )

00

( ) ( ) ( ) ( )

r

r

tX tx x r t x

X

M t E e e x e dx x e dx

rr

−−

1 ( )

0

r t x

x e dx

is finite only if t < λ. Besides, we need to use integration by substitution by

a)

( ) ( )

−− − −

==

==

= = = = − = − = − =

1

100

00

()

( ) ' 1

rrr

rr

X

tt

tt

dM t t r

E X t r t

dt

5.8.9 a)

= = =

− − − −

12

( ) ( ). ( )... ( ) ...

r

r

Y X X X

M t M t M t M t t t t t

5.8.10 a)

= = + +

12

2 2 2 2

12

12

( ) ( ) ( ) exp exp

22

Y X X

tt

M t M t M t t t

Applied Statistics and Probability for Engineers, 7th edition 2017

5-29

Supplemental Exercises

5.S11 The sum of

=

( , ) 1

xy

f x y

,

+ + + + =

1 1 1 1 1 1

4 8 8 4 4

and fXY(x,y) ≥ 0

5.S12 a)

= = = = =

2 4 14

20!

( 2, 4, 14) 0.10 0.20 0.70 0.0631

P X Y Z

Applied Statistics and Probability for Engineers, 7th edition 2017

5-30

5.S13

= = =

3 2 3 23

23

22

00

0 0 0

2 18

23

yx

cx ydydx cx dx c c

. Therefore, c = 1/18.

5.S14 The region x2 + y2 ≤ 1 and 0 < z < 4 is a cylinder of radius 1 (and base area π) and height 4.

Therefore, the volume of the cylinder is 4π and

=1

( , , ) 4

XY Z

f x y z

for x2 + y2 ≤ 1 and 0 < z < 4.

Applied Statistics and Probability for Engineers, 7th edition 2017

5-31

5.S15 Let X, Y, and Z denote the number of problems that result in functional, minor, and no defects,

respectively.

c) E(Z) = 10(0.3) = 3

5.S16 a) Let

X

denote the mean weight of the 25 bricks in the sample. Then,

=( ) 3E X

and

5.S17 Let

X

denote the average time to locate 10 parts. Then,

=4( 5)E X

and

=30

10

X

5.S18 a) Let X denote the weight of a piece of candy and X N(0.1, 0.01). Each package has

Applied Statistics and Probability for Engineers, 7th edition 2017

5-32

5.S19

5.S20 a) Let Y denote the weight of an assembly. Then, E(Y) = 4 + 5.5 + 10 + 8 = 27.5 and V(Y) = 0.42 +

5.S21 Let T denote the total thickness. Then, T = X1 + X2 and

5.S22 Let X and Y denote the percentage returns for security one and two, respectively.

If half of the total dollars is invested in each, then 1/2X + 1/2Y is the percentage return.

5.S23 a) Let X, Y, and Z denote the risk of new competitors as no risk, moderate risk,

Applied Statistics and Probability for Engineers, 7th edition 2017

5-33

5.S24

Y X X Y

==

Then

5.S25

2

I

P I PR

R

==

5.S26 Because the covariance is zero, one may consider a cross section of the beam, say along the

x axis. The distribution along the x axis is

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